Question

Use the Laplace transform to solve the given system of differential equations. frac{(d^2x)}{(dt^2)}+frac{(d^2y)}{(dt^2)}=frac{t}{2} frac{(d^2x)}{(dt^2)}-frac{(d^2y)}{(dt^2)}=4t x(0) = 5, x'(0) = 0, y(0) = 0, y'(0) = 0

Laplace transform
ANSWERED
asked 2021-01-25
Use the Laplace transform to solve the given system of differential equations.
\(\frac{(d^2x)}{(dt^2)}+\frac{(d^2y)}{(dt^2)}=\frac{t}{2}\)
\(\frac{(d^2x)}{(dt^2)}-\frac{(d^2y)}{(dt^2)}=4t\)
\(x(0) = 5, x'(0) = 0,\)
\(y(0) = 0, y'(0) = 0\)

Answers (1)

2021-01-26

Step 1
Given that
\(\frac{(d^2x)}{(dt^2)}+\frac{(d^2y)}{(dt^2)}=\frac{t}{2} \dots(A)\)
\(\frac{(d^2x)}{(dt^2)}-\frac{(d^2y)}{(dt^2)}=4t \dots(B)\)
\(x(0) = 5, x'(0) = 0,\)
\(y(0) = 0, y'(0) = 0\)
adding equation( A) and (B)
\(\Rightarrow 2\frac{(d^2x)}{(dt^2)}=(t^2+4t)\)
\(\frac{(d^2x)}{(dt^2)}=\frac{(t^2+4t)}{2}\)
\(\frac{(d^2x)}{(dt^2)}=\frac{(t^2)}{2}+2t\)
Step 2
taking Laplace transform on both sides,
\(s2X(s)-sx(0)-x'(0)=\frac{1}{2}+ \frac{2}{s^2} + \frac{2}{s^2}\)
\(s2X(s)-5s=\frac{1}{(s^3)}+\frac{2}{s^2}\)
\(s2X(s)=\frac{(1+2s+5s^4)}{s^5}\)
\(X(s)=\frac{1}{s^5}+\frac{2}{s^4}+\frac{5}{s}\)
taking Inverse Laplace transform
\(x(t)=\frac{t^4}{4!}+\frac{(2t^3)}{3!}+5u(t)\)
Step 3
\(u(t)=1, t\geq0\)
\(=0 , t<0\)
\(\therefore x(t)=\frac{(t^4)}{24} + \frac{(t^3)}{3} + 5 \text{ for } t\geq0\)
And Subtracting equation (B) from (A)
\(\Rightarrow 2\frac{(d^2y)}{(dt^2)}=t^2-4t\)
\(\frac{(d^2y)}{(dt^2)}=\frac{(t^2-4t)}{2}=\frac{(t^2)}{2}-2t\)
Taking ILT
\(s^2Y(s)-sy(0)-y'(0)=\frac{1}{2} \frac{2}{s^3}-\frac{2}{s^2}\)
\(s^2Y(s)=\frac{1}{(s^3)}-\frac{2}{(s^2)}=\frac{(1-2s)}{s^3}\)
\(Y(s)=\frac{(1-2s)}{s^5}\)
Step 4
\(Y(s)=\frac{1}{s^5}-\frac{2s}{(s^5)}\)
\(Y(s)=\frac{1}{s^5}-\frac{2}{(s^4)}\)
Taking Inverse Laplace Transform
\(y(t)=\frac{(t^4)}{4!}-\frac{(2t^3)}{3!}\)
\(=\frac{(t^4)}{24}-\frac{(t^3)}{3}\)
\(y(t)=\frac{(t^4)}{24}-\frac{(t^3)}{3}\)

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